Evaluate the following integral using trigonometric substitution.

${\displaystyle \underset{0}{\overset{3\sqrt[\phantom{2}]{2}}{}}}\frac{x^2}{\sqrt[\phantom{2}]{36-x^2}}dx$

A$.x=6tan$

B$.x=6sec$

C$.x=6sin$

Rewrite the given integral using this substitution.

${\displaystyle \underset{0}{\overset{3\sqrt[\phantom{2}]{2}}{}}}\frac{x^2}{\sqrt[\phantom{2}]{36-x^2}}dx={\displaystyle \underset{0}{\overset{\frac{}{4}}{}}}(36si{n}^2)d$

$($Type exact answers.$)$

Evaluate the integral.

${\displaystyle \underset{0}{\overset{3\sqrt[\phantom{2}]{2}}{}}}\frac{x^2}{\sqrt[\phantom{2}]{36-x^2}}dx=,$ dype $an$ exact answer.